Two vessel separately contains two ideal gases A and B at the same temperature, the pressure of A being twice that of B. under such conditions, the density of A is found to be 1.5 times the density of B. the ratio of molecular weight of A and B is

Options

A. `(2)/(3)`
B. `(3)/(4)`
C. 2
D. `(1)/(2)`

Answer

To solve the problem, we will use the ideal gas law and the relationships between pressure, density, and molecular weight of the gases.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Let the pressure of gas A be \( P_A \) and the pressure of gas B be \( P_B \).
- It is given that \( P_A = 2 P_B \).
- Let the density of gas A be \( \rho_A \) and the density of gas B be \( \rho_B \).
- It is given that \( \rho_A = 1.5 \rho_B \).

2. **Use the Ideal Gas Law:**
The ideal gas law is given by:
\[ PV = nRT \]
We can express pressure in terms of density and molecular weight:
\[ P = \frac{\rho}{M} RT \]
where \( \rho \) is the density, \( M \) is the molecular weight, \( R \) is the gas constant, and \( T \) is the temperature.

3. **Express Pressures in Terms of Densities and Molecular Weights:**
For gas A:
\[ P_A = \frac{\rho_A}{M_A} RT \]
For gas B:
\[ P_B = \frac{\rho_B}{M_B} RT \]

4. **Set Up the Ratio of Pressures:**
From the given information:
\[ \frac{P_A}{P_B} = \frac{\rho_A / M_A}{\rho_B / M_B} \]
Substituting the known ratio of pressures:
\[ \frac{2}{1} = \frac{\rho_A}{\rho_B} \cdot \frac{M_B}{M_A} \]

5. **Substitute the Density Ratio:**
We know that \( \rho_A = 1.5 \rho_B \), so:
\[ \frac{2}{1} = \frac{1.5 \rho_B}{\rho_B} \cdot \frac{M_B}{M_A} \]
Simplifying gives:
\[ 2 = 1.5 \cdot \frac{M_B}{M_A} \]

6. **Rearranging to Find the Ratio of Molecular Weights:**
Rearranging the equation:
\[ \frac{M_B}{M_A} = \frac{2}{1.5} = \frac{4}{3} \]
Taking the reciprocal gives:
\[ \frac{M_A}{M_B} = \frac{3}{4} \]

7. **Final Result:**
The ratio of the molecular weights of gases A and B is:
\[ \frac{M_A}{M_B} = \frac{3}{4} \]